Derive Anti Resonance Frequency from Transfer Function I am currently looking for a way to derive the anti-resonance frequency of a transfer function (TF). 
I seems that the resonance frequency can be derived from:
$\left\lvert G(jw) \right\rvert_{max} = \frac{\partial \left\lvert G(jw) \right\rvert}{\partial w} = 0$
since it is the maximum amplitude of the magnitude plot in the bode diagram.
An easy way for an approximation would be to calculate the zeros of the TF (similar to the pole frequencies and their approximation to the resonance freq. which are closely located, see also Resonance from bode plot)
A TF can be for example:
$G(s) = \frac{1}{s^2(J_m+J_l)} \frac{J_l s^2 + ds + c}{\frac{J_m J_l}{J_m+J_l}s^2 + d s + c}$
the bode plot for $J_m = 0.002, J_l = 0.02, d = 0.5, c = 50$ is
Bode Plot
 A: Well, first of all we can write:
$$\left|\text{G}\left(\omega\text{j}\right)\right|=\left|\frac{1}{\left(\omega\text{j}\right)^2\cdot\left(0.002+0.02\right)}\cdot\frac{\left(\omega\text{j}\right)^2\cdot0.02+\omega\text{j}\cdot0.5+50}{\frac{0.002\cdot0.02}{0.002+0.02}\cdot\left(\omega\text{j}\right)^2+\omega\text{j}\cdot0.5+50}\right|=$$
$$\frac{1}{\left|-\omega^2\cdot0.022\right|}\cdot\frac{\left|-\omega^2\cdot0.02+\omega\text{j}\cdot0.5+50\right|}{\left|-\frac{0.00004}{0.022}\cdot\omega^2+\omega\text{j}\cdot0.5+50\right|}=$$
$$\frac{1}{0.022\omega^2}\cdot\frac{\sqrt{\left(50-0.02\omega^2\right)^2+\left(0.5\omega\right)^2}}{\sqrt{\left(50-\frac{0.00004}{0.022}\cdot\omega^2\right)^2+\left(0.5\omega\right)^2}}\tag1$$
Where $\text{j}^2=-1$.
Finding the point in the Bodeplot we need to solve:
$$\frac{\text{d}}{\text{d}\omega}\left(20\log_{10}\left(\left|\text{G}\left(\omega\text{j}\right)\right|\right)\right)=0\space\Longleftrightarrow\space\omega\approx53.8424\space\vee\space\omega\approx106.292\tag2$$
